 Okay. So thanks for all the survivors that are here today. So it's the last talk. And I would like to talk a little bit about the flow case. The flow case from the theorem that we discussed yesterday. So remember that yesterday we considered a two-dimensional, a surface defilmophism that had, well, in general, you assume it has positive topological entropy so that you can have measures that are hyperbolic. And while we were able to code simultaneously all hyperbolic measures, right, by what? By a topological Markov shift. Given an oriented graph, we were considered the space of z-index paths on this graph with the dynamics of the left shift. And that's exactly the structure that we were able to construct as an extension of our defilmophism. So today I would like to consider a three-dimensional flow. It's natural when you have a result in two dimensions for defilmophisms and you want to get four flows. So first consider three dimensions. And that's what we are going to consider. And we are going to assume further that the flow has positive speed. That means that it has no fixed points. So in other words, the vector field that is generated by the flow is non-zero everywhere. And our goal is to try to code, to get a similar coding for hyperbolic measures. So what is the model that we are trying to construct? If you remember in the first lecture I defined what is known as a topological Markov flow. So what is a topological Markov flow? It is the flow version of the topological Markov shift. So it is a suspension flow in a suspension space where the base's dynamics is the dynamics of a shift. And you have a positive roof function that tells you the time it takes for you to start in the section, in the base, and come back to the base. So that is exactly what we aim to construct today. There will be a big difference from the result of Sariq is that we are only able so far to code one measure at a time. So let me say what the setting is. So let M3 be a smooth, closed, compact without boundary, remanian manifold. Let X be a C1 plus beta vector field on M. Positive speed. So I assume that X of p is different from zero everywhere. Let phi be the flow generated by X. And given a positive phi, let mu be a hyperbolic probability measure. So remember that we are in three dimensions and we always have a Lyapunov exponent which is equal to zero. It is in the flow direction. So we have two other Lyapunov exponents and saying that this measure is hyperbolic is saying that almost everywhere one of these two other Lyapunov exponents is bigger than phi and the other is smaller than minus phi. And the theorem that we want to sketch the proof is that under those conditions that exists topological Markov flow, sigma r. This is the suspension space and small sigma r is the suspension flow. And I hold the continuous map phi sub r that comes from sigma r2m such that it satisfies the same three properties that we got for the thermophysm. So the first one is that I'm always confused about that. Pi r composed of sigma rt. The time t of the suspension flow is conjugated by the time t of the flow phi t by pi r. So this happens for all rt's. Also this coding is relevant for the measure mu. So this guy has full mu measure. And finally we have finiteness 2 1. So for every x or I'm just going to write finiteness 2 1. The precise statement is to get a point that is in pi r of sigma r sharp and saying that the number of pre-images of this point in sigma r sharp is finite. But let us forget about that. So note again that I have to fix my measure mu a priori. Given my measure mu, I get a topological Markov flow. If I change my measure mu, possibly I would get another topological Markov flow. So for each measure you have a coding. So what is the idea to prove this theorem? Someone has a suggestion. What do we do when we have flows? Usually we don't like to work with flows, right? We like to work with thermophysm. Exactly. So what I want to do is that I want to mud out the flow direction and try to analyze what happens in the transverse direction. And analyzing what happens in the transverse direction, what does that mean? It means that I will fix a section of my flow. And instead of analyzing the flow, I will analyze the first return map of a point of the section to the section. Okay? So the idea is to mud out the flow direction and analyze the hyperbolicity in the transverse direction. Okay. How? We are going to use Poincare resections and Poincare return maps. Okay? So how do I construct a Poincare resection? So let us say that we have a disk, two-dimensional disk, that is transverse 2x. So I have a small piece of a surface here that I know that in every point the tangent space of the surface does not contain the vector field x. Okay? And I'm just going to introduce a notation now. So given an interval from r, you let phi i of d to be the following. You get d and you flow d with the real times that belong to this interval i. So this is just the union of phi t of d as t belongs to i. Okay? So if this i is non-empty, I'm pretty sure that this set here has non-empty interior. Why? Because my vector field is different from zero everywhere. So what is the picture of this phi i of d? So if I have d like that, I just flow. Let us say that i contains the time zero. So I just flow d a little bit above and possibly a little bit below. And what I get is known as a flow box. So this is phi i of d. And how do I get, how do I construct a Poincare resection? How do I construct a section? So it's a two-dimensional object that sees all the points of my flow? And what do I mean by seeing all the points of my flow? I mean that whenever I get a point of the manifold m, after some finite time, I can flow this point and hit the section. Well, since these guys have non-empty interior, it's very easy to get a Poincare resection by compactness. So by compactness, we can find a finite set of transverse disks, d1 up to dn, such that if you give me an alpha a priori, I can find a point of d. And d i's such that the flow box that you, when you get d i and you flow up to time alpha, when you consider the union of all these flow box, you cover the whole manifold. Okay? And why is that good? That is good because I'm pretty sure that every point of m at most after time alpha is going to hit one of these disks d i. So I call the union of d i's a Poincare resection, lambda equals to the union of d i's. And in order to avoid problems with the return time, let us assume that these d i's are disjoint. So assume finite set of disjoint transverse disks. This might require a little bit of more work. Sometimes you have to get one of the d i's when it intersects a dj. You would have to flow one of the d i's a little bit in order not to intersect dj. But that can be done. Okay? So this is the Poincare resection and what is its property. So for every x in m, and in particular for every x in lambda, there exists a time t of x which belongs to the open interval. It's strictly bigger than zero at most alpha, such that when you get x and you flow by this time t of x, you come back to the section. Okay? So you are starting somewhere here. And then I'm pretty sure that there is a time and this time I'm going to get the minimum, so the minimum x, the minimum positive, such that phi t of x comes back to lambda. So you are here, you flow time, flow the minimum time, such that you come back to lambda. Okay? And I'm going to give a name to this map. The association of x to the position that you first hit the section back is what I'm going to call Poincare Re-Return map. So what is a Poincare Re-Return map? Yeah, I require them to be disjoint here. So what is here? In principle, they are not disjoint, but then I said that you can make them disjoint. By probably whenever you see a non-trivial intersection, you have to flow one of the d i's a little bit in the flow direction. Okay? So that's why I'm pretty sure that the time t is positive. Okay, so the Poincare Re-Return map is this association that I just described to you. It's a map from the section to the section, and f of x is given as the first return of x to the section. Okay? So what is this thing? This thing is a two-dimensional, well, map. Not necessarily a defilmophism. You are going to see soon why it is usually not a defilmophism, but it's two-dimensional. And what we want to see is exactly what we wished to get here. We wanted to mod out the flow direction in order to see hyperbolicity in the transverse direction. So it is exactly with respect to this map that we aim to see non-uniform hyperbolicity. Okay? So how am I going to do that? Well, let x, that belongs to Lambda, that we know that it has a good non-uniform hyperbolicity. So let x, such that there exist vectors e s x and e u x, such that you know that the Lyapunov exponent associated to this first one is smaller than minus e. And while here I have to put the Lyapunov exponent with respect to the flow, so I should divide by t and I should apply the derivative of the time t map of the flow. I assume that this is smaller than minus e, and the other one is bigger than e. t goes to plus minus infinity. Okay? So you have here a surface, or let us say that x belongs to some di, and above x you have the tangent space of x with respect to the manifold, which is three-dimensional, and you have the tangent space with respect to the plate, to the surface, which is two-dimensional. So let us say that this guy here is tx di, and while you know that you have three vectors in the ex m, this, and one of them is the flow direction, you know it is transverse to this plane, and you have two other vectors which give you stable and unstable directions. So this is e s x, and this is e u x. And what I want is to find directions for the point career return map for which I see non-uniform hyperbolicity. So what should I do? I want to mod out the flow direction, right? So what I will do is that, well, if this guy is contracting and if this guy is neutral, I know that this plane here, if you iterate it backwards, it's going to expand. But if I want to mod out this direction, what do I do? I get this plane, and I intersect it with this other plane, I get a direction, and then I just normalize it. So in other words, what I can do is that I can get this e s x and delete and project it into this plane in this direction to get a direction here in tx di. And then I just normalize to get a unit vector. So I'm going to define a vector n s x, which is exactly this. It is the projection of e s x to this plane in this direction. And the same thing with e u. Let me give colors for these guys. And then I get the n u x. So the definition is lambda, actually, unitary, even by the projection e s x to tx lambda in the direction of x. And similarly, okay, and it is in this direction that I wish to see non-uniform hyperbolicity. Well, this is actually true. So you have a lemma, or this is morally true. You need some properties of the angles also, but forget about them, or maybe not. So the lemma is that if you now calculate the Lyapunov exponents of f in this direction, you get something that's negative. And if you calculate the Lyapunov exponent of f in this direction, you get something that's positive. And you can actually give a lower bound of this number, which is the following. So lemma of 1 over n log dfn n s x. This is smaller than minus e prime. It will be a new e. And in the u direction, this is smaller than minus e prime. And you can actually put e prime. It is a multiple of e. And how does it depends on e? It is you can take e prime to be e times the minimum time that you get here in this return map. Minimum of t of x, okay? Well, this number is positive. So what you got is that whenever you see a hyperbolicity bounded away from e, for the map f, you see a hyperbolicity bounded away from e prime, correct? So what is the goal now? Well, the goal is to get the measure that we have in the ambient and consider a new measure here and conclude that this new measure here is going to be e prime hyperbolic. Then we have a two dimensional map, which is, and we have a measure which is e prime hyperbolic and we hope to apply the same methods of the proof of yesterday for f. Okay? Okay, so then if mu is e hyperbolic, we can project mu to a measure nu on lambda. That is e prime hyperbolic. And how do you get this measure nu? The measure nu, some people call it the flux measure. And what is the flux measure? Let us say that you want to measure a subset of your section here. You want to give a reason for this measure here. Okay? So how do you do? You flow this set a little bit above. So you flow it up to time epsilon, for example. And this is now a subset of m. So you can measure it with respect to mu, right? And you want to know what happens in the limit as epsilon goes to zero. So the definition of nu is just, if you call this translation here A epsilon. And actually we know that A epsilon is phi zero epsilon of A. The definition of the flux measure and measures more or less how much we are passing through the section is the limit of the measure mu that we started with. And then you divide by epsilon. Okay? So this measure nu is f invariant by the lemma. It is hip prime hyperbolic. So now what we have is that we have a map that is defined in the surface, two-dimensional guy. And we have a measure that is hip prime hyperbolic. And now everything seems beautiful. We can just apply Sarek's results. Right? Not right. Why? Sarek's results requires that this map f is a defilmophism at least of regularity C1 plus beta. But this map f in general is not even continuous. Why? Because the section lambda is not connected. It is given by the union of these joint disks. So what happens and what will happen is not continuous. And what is the reason of that? It is caused by boundary effects. So what happens if you have this situation here? And you know that this point here hits the section back exactly at the boundary of the next disk. So this boundary here, if you flow it backwards, it defines a curve here. And this curve is separating this disk into two parts. And the property of it is that if you get a point to the left of it, you will hit the same section, the same disk. But if you get a point to the right of it, you will miss this disk. And you will hit another one. So you lose the continuity along this curve. So f is not even continuous. So in order to analyze where it is continuous, let me give a name to the discontinuity set. So let D to be the discontinuity set. And while there is still a hope, and the hope is that, okay, let us see the map f away from this discontinuity set. And let us, okay, outside this discontinuity set, I know that my map f is going to be continuous, and it's actually going to be a local diffeomorphism. If you consider a small neighborhood of the point outside the discontinuity set, well, you can conclude that the map f is a local diffeomorphism onto this small image. Okay, may some? Yes. But that is usually not the case, right? Yeah. Okay, so let D to be the discontinuity set. Well, there is a problem, another problem. What happens if this D is relevant for the dynamics, for the measure? So I want to code almost every point. So I'm aiming to apply the same methods of Sarig. Oh, here's Nu. To code Nu almost every point, right? In particular, I would like this set D to have zero Nu measure. So we want of D to be equal to zero. Otherwise, we will not be able to code all the points, most of the points of our map f. Okay? Okay. So we are going to assume that we have actually a bigger problem. So I'm going to forget for a while about that condition Nu of D equals to zero. And I'm going to try to explain what is the bigger problem. And in order to do that, let us try to follow the steps that we did yesterday in order to construct the symbolic dynamics. So let's try to apply the method or surface defamophisms to the pair or to the triple lambda f Nu. So what is the step one? Step one is to construct passing charts. And actually it was to construct epsilon double charts, but let us first to construct passing charts. How are they defined? They were defined in a domain like this, taking values somewhere given by a composition map, a linear map of the exponential map. Here we can do the same thing. Since we have these two vectors, we can construct that map C of x and we can compose this map C of x that is a map from R2 to the tangent space. We can compose with the exponential map of my section. And that gives me a map from R2 to the section. But note yesterday we defined Q epsilon of x as what? As e to the epsilon to the triple lambda. It's an ugly formula. But what I want to say is that it only depends on the linear behavior of f. It only depends on the derivative of f. Right? Well, the same definition here is not going to work. Because what happens if I have a good hyperbolicity, so this number is big, but my point x is very close to the boundary? Since my point x is very close to the boundary, the exponential map is not defined in this domain. But I cannot define the composition. So in order to avoid that, I should not consider Q of epsilon to be equal to this, but also take into account how far my point x is from the boundary of the section. So I'm going to consider Q of epsilon not as equal to this, but as a minimum of this and the distance of my point x to the boundary of the section. Okay? And now I'm in good shape. I'm able to define the psi x, the passing chart as a map from a subset of R2, a small subset of R2 to my section. So step one is done. Let us try to do step two. Step two, it was coarse-graining. It is done in the same way as before. So I'm just going to write it here and say that it is okay. It can be done. And let us go to step three. And step three is infinite to one extension. Well, I'm going to show you soon. So in principle, you could have a set of positive measures that is converging to the boundary exponentially fast, right? So I want to make sure that the section that I construct does not have this property. And that's the last part of the discussion. So the step three is to construct an infinite to one extension. So remember that the infinite to one extension was a map pi from sigma to m that uses the graph transform method in order to construct this map. So this uses the graph transform method. And remember that in order to apply the graph transform method, we needed the sizes of the charts to be comparable, right? So we applied the graph transform method not to this charts, not because we don't know what is the dependence, how these numbers vary when you take it at x and when you take it at f of x. We needed to pass to a smaller size of charts, which is the small queue of apps that I defined yesterday, and to apply that for the small queue. So this, it only works after we pass to the smaller size of charts. A smaller domain minus q epsilon x, q epsilon x squared. And what was q epsilon x? We defined it to be the minimum of e to the epsilon n, q epsilon fn of x, or integer n. Well, and we want, if we want to get something, we want this q epsilon to be positive, right? This guy is going to be positive. He's saying something about the decay of this. So remember that yesterday I came to you, okay, the q epsilon defined only as this expression, and I know that almost everywhere, this q epsilon doesn't go to zero exponentially fast. And that was the key point to define the q epsilon in order, in a way that it is positive, right? But now I introduced a term here. I no longer know if this q epsilon is going to zero sub-exponentially fast. And I need that, because otherwise the q epsilon that I'm going to define here is going to be zero, and I cannot apply the graph transform method for a point. Okay? So hence, we need that the section lambda satisfies the following. That satisfies that the measure of points that are in lambda, that are approaching the boundary of lambda exponentially fast, which is just this set, I want this measure to be zero. If this measure is zero, I know that for almost every point, this lim if is zero. So I know that these numbers do not go to zero exponentially fast. So I have q epsilon as the minimum of two numbers. Both of them do not go to zero exponentially fast. So I can define the small q like that. Then I have a positive number. Then I can apply the graph transform method and get the infinity to an extension. Okay? So this is the main goal. And I think it answers your question, right? If I can answer this. Yes. So how do we construct a section? So the goal now is to construct a section that satisfies this. A priori, the section that I fixed here in the beginning might not satisfy that. So how do I do? The idea is, okay, I'm not going to consider only one candidate for a section. I'm going to consider many candidates for a section. And I'm actually going to consider one parameter family of candidates for a section. So in order to get this, we consider a one parameter family, lambda t, Poincare sections. And how do I do that? Well, I can just, I can just parametrize them by the radius of each of these disks. So what I'm doing is that I'm fixing lambda and I'm changing a little bit the radii of each of the disks that compose lambda. Okay? Okay, so you can put here parametrized by the radii. Yes. Yes, you increase a little bit, each of them at the same time. No, you get the same number of disks. You just change the radius. So you have many disks here that compose lambda. And what is going to be lambda t? You are going to increase all of them at the same time. Yes. So if you increase them, they will certainly cover m. And if you increase them just a little bit, they will also certainly be disjoint. Okay? They are closed at this. So if you increase a little bit, they will... I don't mind. The cylinders are not supposed to be disjoint. They will never be disjoint. I have a connected manifold, so it cannot be equal to the union of this flow boxes. I just want the di's to be disjoint. So in white here, I have, for example, lambda a. And in red, I have lambda t. Okay, parametrized by the radii of the disks. So what we show is that, well, maybe my initial lambda will not have that property star. But if I consider a one-parameter family of lambdas, then almost every parameter is going to give me the property star. So proposition for lab bag almost every t in a, b. The section lambda t satisfies. How do we do that? Okay. So for each t, let us look at the bad points. Actually, let me do the following. Fix alpha positive and define the bad points for t with quantity alpha. And what is that? It is the set of points in lambda t such that that limit is negative. And not only negative is smaller than or equal to minus alpha. 1 over n log distance f n x boundary of lambda t is smaller than or equal to minus alpha. Okay. Do you agree that if I fix alpha and then I show that for almost every t, this guy has zero measure, then I'm in good shape. Because I can just take an increase in sequence, a decrease in sequence of rational alphas going to zero. And then I intersect all of these sets and I get the result. So my goal is fix this scale alpha and let us show that this guy here has zero measure for almost every parameter t. Okay. We want that the measure of b t is equal to zero for lab bag almost every t in a, b. Okay. So what I will do is the following. b t is the set that you fix t and you look at the bad points x. I will do the opposite. I will fix x and look at the bad parameters t. Okay. So b t is your fix t and look at x. And I'm going to define now a set ex, ex, which is exactly the opposite. You fix x and look at t. So what is that? I x is the parameters t such that when you fix x, you have this property here. So let me give a name for this to be two stars. Okay. So instead of trying to prove, the way that I will show that almost every t gives me zero measure for the set is doing a double counting argument, which is known in the interrogation form as Fubini's theorem. So by Fubini's theorem, I know that if I want to integrate the measure of all the b t's and note that since I want to show that this is zero almost everywhere, it is enough to show that this integral is zero. Right? It's the integral of no negative number. So if it's zero, it's equivalent of the, to this b zero almost everywhere. So how do I calculate this integral? I change the order of integration and I calculate the lab bag measure or the size of the eyes of x. Fubini. Okay. Now I'm, what I want to show is that this right hand side, this right hand integral is equal to zero. Actually, we have that the integrant function is identically equal to zero. Claim the lab bag measure of all of these intervals is zero. Why? What that means for a point to be here? You are fixing an x. So you have the itinerary of x in the section. For a point to be here, it needs to define a radius such that the boundary of this section is exponentially close to the trajectory of this point x. So if you have the trajectory, it means that the boundary of the section has to be in the limb soup of some intervals that whose sizes are going to zero exponentially fast. So what do I mean? I mean that I of x is the limb soup of some intervals I n, where I n is an interval of r of size approximately e to the minus alpha n. And what do I mean by that? That happens because if you have this, then it means that you have this picture here. So you have the center of the disk. Let me do it bigger. You have the center of the disk here. And you have the point f n of x. If this distance is smaller than this, it means that the section that you are considering here, the time t, should define a section for which this distance here is of the order of smaller than or equal to e to the minus alpha n. So that interval is the interval that is centered at this distance, the distance of the center of the disk to f n of x, and whose radius is e to the minus alpha n. So this is the interval I n. And what is the property of the interval I n? It is exponentially small. So in particular, the sum of its lengths is smaller than 0, smaller than infinity, right? So by Borrel Cantelli, you know that the limb soup of these intervals, and what is the limb soup, is the set of points that belong to infinitely many of these intervals. You know that the Lebesgue measure of this limb soup is 0, and that's the end of the proof. So since the sum of I n is smaller than infinity, Borrel Cantelli implies to you that the Lebesgue measure of I of x is 0. And you just showed that almost every parameter t defines to you a section that satisfies this star property. Since it satisfies this star property, we can apply the step three. We can get an infinite one extension. And then we can apply the step four, which is the Bowen-Snag refinement, and get the coding for f. Once you get the coding for f, it is very easy to get the coding for the flow itself. You just suspend the coding for f by the time it takes for you to start in the section and come back to the section, and you are done. Okay? Question or observation. Why can't we do all the same time for all the measures? Why can't we do that for all the measures? Well, we could, if we had this for all the measures, all the flux measures coming from hyperbolic measures. Well, our methods, you see, you have to fix the measure in order to apply, for example, this Fubini argument here. So we are not able to code all the measures simultaneously. Okay? And thank you for your time. So this is the end of my mini course.